The formula for eccentricity, e, of an orbit is given below, where a is the length of the semi-major axis and b is the length of the major axis.

Solve the formula for eccentricity for the length of the semi-major axis.

The correct equation for the length of the semi-major axis is equation .

When the length of the major axis is 4 inches and the eccentricity of the orbit is , the length of the semi-major axis is inches (round to the nearest hundredth).

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Part 1)

Part 2)

Step-by-step explanation:

The correct question in the attached figure

Part 1)

we have the formula

where

e the eccentricity

a is the length of the semi-major axis

b is the length of the major axis

solve for a

That means

Isolate the variable a

squared both sides

subtract 1 both sides

Multiply both sides by a^2

Divide both sides by (e^2-1)

Rewrite

square root both sides

simplify

Part 2) we have

Solve for a

substitute in the formula of part 1)

∠lno ≅ ∠lnm

∠oln ≅ ∠mln

step-by-step explanation:

given   that triangle lno is congruent to triangle lnm.

we have to fill up the blanks in the given question.

according to above order of congruency   kept in mind we can say that

∠lno ≅ ∠lnm

∠oln ≅ ∠mln

ln ≅ ln due to common in both.

hence, by sas congruency,

the side angle side rule   states that if two sides and the included angle of one triangle are congruent to another triangle, then these two triangles are congruent.

c) the median for town a 30" is greater than the median for town b 25"

step-by-step explanation:

town a, median = q2 = 30

town b, median - q2 = 25

so the median for town a 30" is greater than the median for town b 25"